Question

How do you solve `y=1/2x^2-4, y=3x+4` using substitution?

Answer 1

Simply substitute the second equation into the first (both are solved for `y` already).

`3x + 4= 1/2x^2 - 4`

Multiply by a factor of `2` so that all terms are integers.

`6x + 8 = x^2 - 8`

`0 = x^2 - 6x - 16`

`0 = (x - 8)(x+ 2)`

`x = 8 and -2`

This will mean that `y = 3(8) + 4 = 28` and `y = 3(-2) + 4 = -2`

Therefore, the intersection points are `(8, 28)` and `(-2, -2)`.

Hopefully this helps!